$\frac{4y + 24}{5y^2 - 45} + \frac{y + 3}{5y^2 - 15y} = \frac{y - 3}{y^2 + 3y}$
$\frac{4y + 24}{5(y^2 - 9)} + \frac{y + 3}{5y(y - 3)} - \frac{y - 3}{y(y + 3)} = 0$
$\frac{4y + 24}{5(y - 3)(y + 3)} + \frac{y + 3}{5y(y - 3)} - \frac{y - 3}{y(y + 3)} = 0$
$\frac{y(4y + 24) + (y + 3)^2 - 5(y - 3)^2}{5y(y - 3)(y + 3)} = 0$
$\frac{4y^2 + 24y + y^2 + 6y + 9 - 5(y^2 - 6y + 9)}{5y(y - 3)(y + 3)} = 0$
$\frac{4y^2 + 24y + y^2 + 6y + 9 - 5y^2 + 30y - 45}{5y(y - 3)(y + 3)} = 0$
$\frac{60y - 36}{5y(y - 3)(y + 3)} = 0$
$\begin{equation*} \begin{cases} 5y ≠ 0 &\\ y - 3 ≠ 0 &\\ y + 3 ≠ 0 &\\ 60y - 36 = 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} y ≠ 0 &\\ y ≠ 3 &\\ y ≠ -3 &\\ 60y = 36 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} y ≠ 0 &\\ y ≠ 3 &\\ y ≠ -3 &\\ y = \frac{36}{60} & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} y ≠ 0 &\\ y ≠ 3 &\\ y ≠ -3 &\\ y = \frac{3}{5} & \end{cases} \end{equation*}$
Ответ: $y = \frac{3}{5}$

$\frac{y + 2}{8y^3 + 1} - \frac{1}{4y + 2} = \frac{y + 3}{8y^2 - 4y + 2}$
$\frac{y + 2}{8y^3 + 1} - \frac{1}{4y + 2} - \frac{y + 3}{8y^2 - 4y + 2} = 0$
$\frac{y + 2}{(2y + 1)(4y^2 - 2y + 1)} - \frac{1}{2(2y + 1)} - \frac{y + 3}{2(4y^2 - 2y + 1)} = 0$
$\frac{2(y + 2) - (4y^2 - 2y + 1) - (2y + 1)(y + 3)}{2(2y + 1)(4y^2 - 2y + 1)} = 0$
$\frac{2y + 4 - 4y^2 + 2y - 1 - (2y^2 + y + 6y + 3)}{2(2y + 1)(4y^2 - 2y + 1)} = 0$
$\frac{4y + 3 - 4y^2 - 2y^2 - y - 6y - 3}{2(2y + 1)(4y^2 - 2y + 1)} = 0$
$\frac{-6y^2 - 3y}{2(2y + 1)(4y^2 - 2y + 1)} = 0$
$\frac{-3y(2y + 1)}{2(2y + 1)(4y^2 - 2y + 1)} = 0$
$\frac{-3y}{2(4y^2 - 2y + 1)} = 0$
$\begin{equation*} \begin{cases} 2(4y^2 - 2y + 1) ≠ 0 &\\ -3y = 0 & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} 4y^2 - 2y + 1≠ 0 &\\ y = 0 & \end{cases} \end{equation*}$
Ответ: y = 0